用Mathematica作曲線擬合舉例

1、六、用 Mathematica作曲線擬合舉例1 .用三次多項(xiàng)式對(duì)數(shù)據(jù)表1.001,1 ,1.9,4 , 3,9.2 4,15.9進(jìn)行曲線擬 合.d1 1.001,1 ,1.9, 4 ,3, 9.2 ,4, 15.9 TOC o 1-5 h z 1.0011I 1.949.215.9Fit d1,1, x, xA2,xA3, x0.0813373x3 0.215492x2 2.18207x 1.481762.對(duì)數(shù)據(jù)表1.001,1,1.9,4,3,9.2,4,15.9,(1)畫散點(diǎn)圖；(2)用二次多 項(xiàng)式進(jìn)行曲線擬合；(3)畫出擬合曲線；(4)將散點(diǎn)圖和擬合曲線放在一張圖 上觀察擬合效果.d1

2、1.001,1 ,1.9,4 ,3, 9.2 ,4, 15.91.0011I 1.949.215.9st ListPlotd1, PlotStylePointSize.021614 -12：10:8：6；4；1.522.533.54Graphicsnh Fit d1,1, x, xA2, x0.82555x2 0.82101x 0.618748nt Plot nh, x, 1, 4GraphicsShow st, ntGraphics3.對(duì)數(shù)據(jù)表1.001,1,1.9,4,3,9.2,4,15。,(1)畫散點(diǎn)圖；(2)用一次多項(xiàng)式進(jìn)行曲線擬合；(3)畫出擬合曲線；(4)將散點(diǎn)圖和擬合曲線放在一

3、張圖上觀察擬合效果. TOC o 1-5 h z d11.001,1,1.9,4,3,9.2 ,4,15.91.00111.949.215.9.03st3 ListPlot d1, PlotStyle Hue 0 , PointSize16 114 -*io： 8 -6；4：$1.522.533.54Graphicsnh3 Fit di, 1, x , x4.95557x 4.741282345GraphicsShow st3, nt3Graphics.當(dāng)x=1,2,3,4,5,6時(shí)造函數(shù)sinx的函數(shù)表，用三次多項(xiàng)式進(jìn)行曲線擬合.data4 N Table Sin x , x, 1,60.8

4、41471,0.909297,0.14112, 0.756802, 0.958924, 0.279415Fit data4, 1, x, xA3,xA5, x0.00108081x5 0.0471705x3 0.114681x 0.850606.當(dāng)x =0.2,0.4,0.6,0.8,1時(shí),造函數(shù)In sinx的函數(shù)表,(1)畫散點(diǎn)圖；(2)將散 點(diǎn)圖連點(diǎn)成折線；(3)以sin x,sin 2x為基函數(shù)對(duì)該數(shù)據(jù)表進(jìn)行曲線擬合 ;(4) 畫出擬合函數(shù)；(5)將散點(diǎn)圖，折線和擬合曲線放在一張圖上.data5 Table x, Log Sin x , x, 0.2,1., 0.20.21.61611

5、0.40.9431010.60.5715630.80.3321831.0.172604nh5 Fit data5, 1, Sin x , Sin 2 x , xPointSize1.39652sin x0.940588sin 2 x2.22709st5 ListPlot data5, PlotStyleHue .7.03-0.2 -0.4 -0.6-0.80.20.40.60.8Graphicszx5=ListPlotdata5, PlotJoined - True-1.2-1.4-1.6 IGraphicsnt5 Plot nh5, x, 0, 1 , PlotStyle Hue .9-0.

6、20.40.60.81- -0.5-1 -1.5 Z , -2 -GraphicsShow st5, zx5, nt5Graphics6.為了測(cè)定刀具的磨損速度，我們這樣做的實(shí)驗(yàn)：經(jīng)過(guò)一定的時(shí)間t（如每隔一小時(shí)）,測(cè)量一次刀具的厚度y,得到一組實(shí)測(cè)數(shù)據(jù)如下：t01234567y27.26.826.526.326.125.725.324.8試根據(jù)上面的實(shí)驗(yàn)數(shù)據(jù)用最小二乘法建立y和t之間的經(jīng)驗(yàn)公式y(tǒng)=f（t）.也就是說(shuō)，要找出一個(gè)能使上述數(shù)據(jù)大體適合的函數(shù)關(guān)系.第一步畫出數(shù)據(jù)表的散點(diǎn)圖，以確定f的類型.（*輸入數(shù)據(jù)*）d 0, 27.0,1, 26.8,2, 26.5,3, 26.3,4, 26.

7、15, 25.7,6, 25.3,7, 24.8;（*作出數(shù)據(jù)表的散點(diǎn)圖*）st ListPlotd, PlotStyleHue 0 , PointSize.0327 26.52625.5Graphics由圖可見(jiàn)，數(shù)據(jù)點(diǎn)大致在一條直線附近.可以認(rèn)為f是線性函數(shù).因此設(shè)f(t尸at+b第2步 構(gòu)造偏差平方和Q a_, bd i, 227. b 27 a b 24.826a b 25.325 a b 25.724a b 26.1 23 a b 26.322 a b 26.52a b 26.82第3步求出待定參數(shù)a, bSolvea Q a, b0, b Q a, b0 , a, b0.303571

8、,b27.125第4步寫出擬合函數(shù)f t 0.303571 t 27.1257.用Fit對(duì)6題中的數(shù)據(jù)表分別進(jìn)行直線擬合和拋物擬合，并比較那種擬合更好？26.1（*輸入數(shù)據(jù)*） TOC o 1-5 h z d 0, 27.0,1, 26.8,2, 26.5,3, 26.3,4,5, 25.7,6, 25.3,7, 24.8;（*作出數(shù)據(jù)表的散點(diǎn)圖*）st ListPlot d, PlotStyle Hue 0 , PointSize .0327 f26.5 -* 26 -25.5 1123456$Graphics（*直接用Mathematica求出直線擬合函數(shù)*）nh t_ Fit d, 1,

9、 t , t27.125 0.303571t（*直接用Mathematica求出拋物擬合函數(shù)*）nhp t_ Fit d, 1, t, t A 2 , t0.0232143t20.141071t 26.9625（*計(jì)算直線擬合nh的偏差平方和*）q18d k, 2 nh d k, 12k 10.108214（*計(jì)算拋物擬合nhp的偏差平方和*）q28d k, 2 nhp d k, 12k 10.0176786If q1 q2, Print 直線擬合好 ,Print 拋物擬合好拋物擬合好（*作出直線擬合函數(shù)的圖形*）nt Plot nh tt, 0, 8Graphics（*作出拋物擬合函數(shù)的圖形

10、*）ntp Plot nhp t , t, 0, 8Graphics（*在同一坐標(biāo)系中顯示散點(diǎn)圖和擬合曲線的圖形*）Show st, nt, ntpGraphics8.對(duì)6題中的數(shù)據(jù)表分別進(jìn)行指數(shù)擬合和拋物擬合，并比較那種擬合更好？（*輸入數(shù)據(jù)*）d 0, 27.0,1, 26.8,2, 26.5,3, 26.3,4, 26.15, 25.7,6, 25.3,7, 24.8;（*直接用Mathematica求出拋物擬合函數(shù)*）nhp t_ Fit d, 1, t, t A 2 , t0.0232143t20.141071t 26.9625（*計(jì)算拋物擬合nhp的偏差平方和*）qi8d k, 2 nhp d k, 1 k 10.0176786StatisticsNonlinearFit（*直接用Mathematica求出指數(shù)